Tim plans to attempt exam until pass. Chance of passing each time is $20\%$. Find probability that he needs to attempt atleast $2$ & atmost $5$ times Question : Tim is in no hurry to graduate. He intends to write his last exam until he passes it, but he is too lazy to study. He estimates that the chance of passing the exam each time is $20\%$. Determine the probability that he needs to write the exam at least $2$ and at most $5$ times.
Correct Answer : $0.472$
My Attempt

Let $p$ be probability of success. Hence $\boxed{p=0.2}$
Let $n$ be number of trials. Hence $\boxed{n=5}$
$X \in Binomial(5, 0.2)$

Hence,
$P(X \ge 2) \cdot P(X \le 5) = 0.26272$

I don't know what I am doing wrong here, I am really grateful for any help or guidance :)
 A: Alternative approach:
You are looking for the probability of the union of the mutually exclusive events $E_k ~: k \in \{1,2,3,4\}$, where $E_k$ represents the event that the test is failed exactly $k$ times, and then passed.
$p(E_k) = (0.8)^k \times (0.2).$
Since these events are mutually exclusive, you can obtain the probability of their union by summing each event.
Thefore, the desired probability is
$$(0.8)(0.2) + (0.8)^2(0.2) + (0.8)^3(0.2) + (0.8)^4(0.2)$$
$$= (0.8)(0.2) [~1 + (0.8) + (0.8)^2 + (0.8)^3]$$
$$= (0.8)(0.2) \times \frac{1 - (0.8)^4}{1 - 0.8}$$
$$ = (0.8) \times \left[1 - (0.8)^4\right] = (0.8) - (0.8)^5 = 0.47232. \tag1 $$

I performed the computation in the above (somewhat convoluted) manner intentionally, to stretch your intuition.  Notice the LHS of (1) above.
This result may alternatively be immediately intuited by reasoning that Tim must fail the first exam, but must not fail all of the next $4$ consecutive exams.
This intuits $(0.8) \times [1 - (0.8)^4].$
A: The probability that he passes on his second try should be modelled by a geometric distribution X~Geo(0.2) [ I guess you could use negative-binomial too]
P(X=r) would be 0.2(0.8)^(r-1)
Now for P(2≤X≤5), You would have to sum the values of P(X=2) to P(X=5) which would give you 0.472
